What is the extreme f(x)=(sqrt(2x-x^2))/x ?

The extreme f(x)=(sqrt(2x-x^2))/x is Minimum(2,0)

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extreme f(x)=(sqrt(2x-x^2))/x

extreme points

f (x) = \frac{2 x - x ^{2}}{x}

Minimum (2, 0)

Solution steps

f^{'} (x) = - \frac{1}{x - x ^{2} + 2 x}

Find intervals:

Decreasing

: 0 < x < 2

Plug

x = 2

into

\frac{2 x - x ^{2}}{x} : 0

Minimum (2, 0)

f (x, y) = 2 x^{3} - y^{3} - 24 x + 27 y - 2

f (x) = (x + 2)^{3}

f (x) = \frac{4 x}{x ^{2} - 36}

f (x, y) = 1 - e^{- x - y}

3 x^{4} - 4 x^{3} - 12 x^{2}

What is the extreme f(x)=(sqrt(2x-x^2))/x ?
The extreme f(x)=(sqrt(2x-x^2))/x is Minimum(2,0)